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denis-greek [22]
3 years ago
15

Please answer This Just Try Don't skip​

Mathematics
2 answers:
Sav [38]3 years ago
7 0
(i)
Area of rectangle = length * breadth
So,
x^2 - 12x + 35 = (x - 5) * breadth

Factorize area and divide by length to simplify.
=> {(x-5)(x-7)} / (x-5) = breadth
=> x-7 = breadth

Therefore, breadth = 7

(ii)
Length = 10 - 5 = 5m
Breadth = 10 - 7 = 3m
Area = 10^2 - 12(10) + 35 = 15sq.m
attashe74 [19]3 years ago
4 0

Answer:

5 m , 3 m , 15 m²

Step-by-step explanation:

Given area (A) = x² - 12x + 35 and A = length × breadth , then

x² - 12x + 35 = (x - 5)(x - 7) ← in factored form

We know length = x - 5 , then breadth = x - 7

When x = 10

length = x - 5 = 10 - 5 = 5 m and breadth = x - 7 = 10 - 7 = 3 m

A = 10² - 12(10) + 35 = 100 - 120 + 35 = 15 m²

or

A = length × breadth = 5 × 3 = 15 m²

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Consider the following equations. f(x) = − 4/ x3, y = 0, x = −2, x = −1. Sketch the region bounded by the graphs of the equation
Sindrei [870]

Answer:

1.5 unit^2

Step-by-step explanation:

Solution:-

- A graphing utility was used to plot the following equations:

                         f ( x ) = - \frac{4}{x^3}\\\\y = 0 , x = -1 , x = -2

- The plot is given in the document attached.

- We are to determine the area bounded by the above function f ( x ) subjected boundary equations ( y = 0 , x = -1 , x = - 2 ).

- We will utilize the double integral formulations to determine the area bounded by f ( x ) and boundary equations.

We will first perform integration in the y-direction ( dy ) which has a lower bounded of ( a = y = 0 ) and an upper bound of the function ( b = f ( x ) ) itself. Next we will proceed by integrating with respect to ( dx ) with lower limit defined by the boundary equation ( c = x = -2 ) and upper bound ( d = x = - 1 ).

The double integration formulation can be written as:

                           A= \int\limits_c^d \int\limits_a^b {} \, dy.dx \\\\A = \int\limits_c^d { - \frac{4}{x^3} } . dx\\\\A = \frac{2}{x^2} |\limits_-_2^-^1\\\\A = \frac{2}{1} - \frac{2}{4} \\\\A = \frac{3}{2} unit^2

Answer: 1.5 unit^2 is the amount of area bounded by the given curve f ( x ) and the boundary equations.

Download docx
3 0
3 years ago
What is the maximum number of rides Karen can go on? A) 14 B) 16 C) 18 D) 20​
Nataly_w [17]

Answer:

I sorry to say but there's not enough information to go of of

4 0
2 years ago
Subtract -2x^2+4x-1−2x
vivado [14]

Answer:

8x²- x -8

Step-by-step explanation:

The question is not well written. I guess what you mean is:

Subtract -2x^2+4x-1 ( minus 2x squared plus 4x minus 1) from 6x^2+3x-9 (6x squared plus 3x minus 9).

To  subtract -2x^2+4x-1 from 6x^2+3x-9, that is

6x^2+3x-9 - (-2x^2+4x-1). This can be written as

6x²+3x-9 - (-2x²+4x-1)

Now, opening the bracket, we will get

6x²+3x-9 +2x²-4x+1

Then, collecting like terms, we get

6x²+2x²+3x-4x-9+1

6x²+2x² = 8x²

3x-4x = -x

-9+1 = -8

∴6x²+2x²+3x-4x-9+1 becomes

8x²-x -8

The answer is 8x²-x -8.

7 0
3 years ago
Sin4x.sin5x+sin4x.sin3x-sin2x.sinx=0
andreev551 [17]

Recall the angle sum identity for cosine:

cos(<em>x</em> + <em>y</em>) = cos(<em>x</em>) cos(<em>y</em>) - sin(<em>x</em>) sin(<em>y</em>)

cos(<em>x</em> - <em>y</em>) = cos(<em>x</em>) cos(<em>y</em>) + sin(<em>x</em>) sin(<em>y</em>)

==>   sin(<em>x</em>) sin(<em>y</em>) = 1/2 (cos(<em>x</em> - <em>y</em>) - cos(<em>x</em> + <em>y</em>))

Then rewrite the equation as

sin(4<em>x</em>) sin(5<em>x</em>) + sin(4<em>x</em>) sin(3<em>x</em>) - sin(2<em>x</em>) sin(<em>x</em>) = 0

1/2 (cos(-<em>x</em>) - cos(9<em>x</em>)) + 1/2 (cos(<em>x</em>) - cos(7<em>x</em>)) - 1/2 (cos(<em>x</em>) - cos(3<em>x</em>)) = 0

1/2 (cos(9<em>x</em>) - cos(<em>x</em>)) + 1/2 (cos(7<em>x</em>) - cos(3<em>x</em>)) = 0

sin(5<em>x</em>) sin(-4<em>x</em>) + sin(5<em>x</em>) sin(-2<em>x</em>) = 0

-sin(5<em>x</em>) (sin(4<em>x</em>) + sin(2<em>x</em>)) = 0

sin(5<em>x</em>) (sin(4<em>x</em>) + sin(2<em>x</em>)) = 0

Recall the double angle identity for sine:

sin(2<em>x</em>) = 2 sin(<em>x</em>) cos(<em>x</em>)

Rewrite the equation again as

sin(5<em>x</em>) (2 sin(2<em>x</em>) cos(2<em>x</em>) + sin(2<em>x</em>)) = 0

sin(5<em>x</em>) sin(2<em>x</em>) (2 cos(2<em>x</em>) + 1) = 0

sin(5<em>x</em>) = 0   <u>or</u>   sin(2<em>x</em>) = 0   <u>or</u>   2 cos(2<em>x</em>) + 1 = 0

sin(5<em>x</em>) = 0   <u>or</u>   sin(2<em>x</em>) = 0   <u>or</u>   cos(2<em>x</em>) = -1/2

sin(5<em>x</em>) = 0   ==>   5<em>x</em> = arcsin(0) + 2<em>nπ</em>   <u>or</u>   5<em>x</em> = arcsin(0) + <em>π</em> + 2<em>nπ</em>

… … … … …   ==>   5<em>x</em> = 2<em>nπ</em>   <u>or</u>   5<em>x</em> = (2<em>n</em> + 1)<em>π</em>

… … … … …   ==>   <em>x</em> = 2<em>nπ</em>/5   <u>or</u>   <em>x</em> = (2<em>n</em> + 1)<em>π</em>/5

sin(2<em>x</em>) = 0   ==>   2<em>x</em> = arcsin(0) + 2<em>nπ</em>   <u>or</u>   2<em>x</em> = arcsin(0) + <em>π</em> + 2<em>nπ</em>

… … … … …   ==>   2<em>x</em> = 2<em>nπ</em>   <u>or</u>   2<em>x</em> = (2<em>n</em> + 1)<em>π</em>

… … … … …   ==>   <em>x</em> = <em>nπ</em>   <u>or</u>   <em>x</em> = (2<em>n</em> + 1)<em>π</em>/2

cos(2<em>x</em>) = -1/2   ==>   2<em>x</em> = arccos(-1/2) + 2<em>nπ</em>   <u>or</u>   2<em>x</em> = -arccos(-1/2) + 2<em>nπ</em>

… … … … … …    ==>   2<em>x</em> = 2<em>π</em>/3 + 2<em>nπ</em>   <u>or</u>   2<em>x</em> = -2<em>π</em>/3 + 2<em>nπ</em>

… … … … … …    ==>   <em>x</em> = <em>π</em>/3 + <em>nπ</em>   <u>or</u>   <em>x</em> = -<em>π</em>/3 + <em>nπ</em>

<em />

(where <em>n</em> is any integer)

5 0
3 years ago
PLEASEEE HELPPPPPPP :))))))))
Oxana [17]

Answer:

False

Explanation

The x-intercept of a line occurs when the line intersects with the x-axis.

8 0
2 years ago
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